Question

Household   Paper   Plastic
1   11.36   10.25
2   7.72   3.86
3   13.05   12.31
4   17.65   11.26
5   6.33   3.86
6   11.42   12.81
7   8.82   11.89
8   9.41   3.36
9   7.98   6.09
10   6.96   7.60
11   8.72   9.20
12   6.05   2.73
13   16.08   14.36
14   6.98   2.65
15   6.16   5.88
16   14.33   6.43
17   13.61   8.95
18   6.44   8.40
19   9.45   3.02
20   12.43   8.57
21   9.83   6.26
22   2.80   5.92
23   12.73   14.83
24   20.12   18.35
25   9.55   9.20
26   15.09   9.11
27   16.39   9.70
28   5.86   3.91
29   6.83   3.57
30   13.31   19.70

Refer to the data set in the accompanying table. Assume that the paired sample data is a simple random sample and the differences have a distribution that is approximately normal. Use a significance level of

0.010.01

to test for a difference between the weights of discarded paper​ (in pounds) and weights of discarded plastic​ (in pounds).

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Click the icon to view the data.In this​ example,

mu Subscript dμd

is the mean value of the differences d for the population of all pairs of​ data, where each individual difference d is defined as the weight of discarded paper minus the weight of discarded plastic for a household. What are the null and alternative hypotheses for the hypothesis​ test?

A.

Upper H 0H0​:

mu Subscript dμdequals=0

Upper H 1H1​:

mu Subscript dμdnot equals≠0

B.

Upper H 0H0​:

mu Subscript dμdnot equals≠0

Upper H 1H1​:

mu Subscript dμdgreater than>0

C.

Upper H 0H0​:

mu Subscript dμdnot equals≠0

Upper H 1H1​:

mu Subscript dμdequals=0

D.

Upper H 0H0​:

mu Subscript dμdequals=0

Upper H 1H1​:

mu Subscript dμdless than<0

Identify the test statistic.

T STAT Equals=

​(Round to two decimal places as​ needed.)

Identify the​ P-value.

​P-value is ___

​(Round to three decimal places as​ needed.)

What is the conclusion based on the hypothesis​ test?

Since the​ P-value is

greater / less than the significance​ level,

fail to reject OR

reject

the null hypothesis. There

is OR

is not

sufficient evidence to support the claim that there is a difference between the weights of discarded paper and discarded plastic

Answer 1

MS Excel was used to claculated mean of Sample difference and Stadnard Deviation of sample difference. screenshot below:

The null and alternative hypotheses for the hypothesis​ test are as follows (Option A):

H0​: μd = 0

H1​: μd ≠ 0

Test Statstic t = d̅/(sd/sqrt(n))

Substituting d̅ = 1.9810, sd = 3.4291 and n = 30, we get:

Test Statstic t = 1.9810/(3.4291/sqrt(30))

Test Statstic t = 3.16

Degrees of Freedom = n - 1 = 30 - 1 = 29

P-value corrsponding t = 3.16 and df = 29 for a two tailed test is obtianed using p-value calculator

P-value = 0.004

Since p-value = 0.004 is less than 0.01 significance level, we reject null hypothesis

There is sufficient evidence to support the claim that there is a difference between the weights of discarded paper and discarded plastic